Mathematics 2016 Jamb Examination

First, perform the multiplication in the numerator: \[ 0.026 \times 0.36 = 0.00936 \] Next, divide by the denominator: \[ \dfrac{0.00936}{0.69} \approx 0.013565... \] Rounding to three significant figures gives \( 0.0136 \). In standard form (scientific notation): \[ 1.36 \times 10^{-2} \]

The ratio is 2:3:5. Let the total number of pencils be \( x \). Bisi's share is represented by 2 parts out of the total 10 parts \( (2+3+5) \). \[ \dfrac{2}{10} \times x = 5 \] \[ \dfrac{1}{5}x = 5 \] \[ x = 5 \times 5 = 25 \] Total pencils shared were 25.

The perimeter of a sector is the sum of two radii and the arc length: \[ \text{Perimeter} = 2r + \left( \dfrac{\theta}{360} \times 2\pi r \right) \] Substituting \( r = 9 \) and \( \theta = 36 \): \[ \text{Perimeter} = 2(9) + \left( \dfrac{36}{360} \times 2\pi(9) \right) \] \[ \text{Perimeter} = 18 + \left( \dfrac{1}{10} \times 18\pi \right) = 18 + \dfrac{9\pi}{5} \; \text{cm} \]

Simplify each part using indices rules: \[ 27^{1/3} = \sqrt[3]{27} = 3 \] \[ 8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4 \] \[ 16^{2/4} = 16^{1/2} = \sqrt{16} = 4 \] Substitute back into the expression: \[ \dfrac{3 - 4}{4 \times 2} = \dfrac{-1}{8} \]

First, arrange the scores in ascending order with their frequencies: Score 2 (Freq 6), Score 3 (Freq 2), Score 5 (Freq 4), Score 6 (Freq 3). Total frequency \( N = 15 \). The median is the \( \frac{15+1}{2} = 8^{\text{th}} \) value. Cumulative frequencies: 6, 8, 12, 15. The \( 8^{\text{th}} \) value falls in the 'Score 3' group.

Rewrite the function: \( y = \dfrac{2}{3}x^3 - 4x^{-1} \). Differentiate using the power rule: \[ \dfrac{dy}{dx} = \dfrac{2}{3}(3x^2) - 4(-1x^{-2}) \] \[ \dfrac{dy}{dx} = 2x^2 + 4x^{-2} = 2x^2 + \dfrac{4}{x^2} \]

Calculate numerator and denominator: \[ \text{Numerator} = 0.010664 \] \[ \text{Denominator} = 0.000756 \] \[ \dfrac{0.010664}{0.000756} \approx 14.10582... \] To four significant figures, we get 14.11.

Use the formula for arc length \( L \): \[ L = \dfrac{\theta}{360} \times 2\pi r \] \[ L = \dfrac{30}{360} \times 2\pi(12) = \dfrac{1}{12} \times 24\pi = 2\pi \; \text{cm} \]

Integrate \( \sin x \) to get \( -\cos x \): \[ \int_{0}^{\pi/2} \sin x dx = [-\cos x]_{0}^{\pi/2} \] \[ = -(\cos(\pi/2) - \cos(0)) \] \[ = -(0 - 1) = 1 \]

Using the arc length formula: \[ L = \dfrac{\theta}{360} \times 2\pi r \] \[ 16\pi = \dfrac{80}{360} \times 2\pi r \] \[ 16 = \dfrac{160}{360}r \Rightarrow 1 = \dfrac{10}{360}r \] \[ r = \dfrac{360}{10} = 36 \; \text{cm} \]